Friction/acceleration of cylinder

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SUMMARY

The discussion centers on the dynamics of a spool being pulled by a tensioned string, where the spool has a mass 'm' and a rotational inertia defined as I=(1/2)mr^2. Participants concluded that the spool will experience angular acceleration in a clockwise direction while linear acceleration moves to the right, with friction acting to the left. The key takeaway is that the direction of motion depends on the balance of forces, particularly the tension and friction, with the condition of no slip at the contact point being crucial to the analysis.

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Homework Statement


given a situation (http://teachingphysics.files.wordpress.com/2009/03/spool-21.png" ) where a string with tension is pulling a spool of mass m from the bottom, what would be the directions of acceleration and friction? note: there's no slip at all.

given: tension equals weight (mg). the spool has rotational inertia I=(1/2)mr^2


Homework Equations


(not trying to calculate anything)


The Attempt at a Solution


the spool would wind up, so velocity would be to the right I think. if you make the axis at where it touches the ground, tension is the only force providing torque...so I think angular acceleration would be clockwise, and so linear acceleration would be pointing right, and friction would be towards the left?

I'm not really sure, so any help would be appreciated.
 
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It sure looks to me like pulling the string will try to unwind the spool and make it turn counterclockwise so it will try to go left. Friction will give it a grip on the table. You'll have a left force AND a right force which will have to be calculated to see which way it goes. It may well go right and wind the string if the friction is sufficiently strong.
 
hint: since there is no slip acceleration at point of contact is zero
 

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